s/(\d[,.]\d)|([^\w\s](?!\s))/$1 || "$2 "/ge;
[download]

Abigail

Thank you, Abigail. Can you please explain how does the replacement part of your expression work? If I understand correctly, /e evaluates $1 || "$2 " -- that way if $1 is not empty, it should not add second part?

--dda

P.S. I added end-of-line processings as was suggested by aragorn:

s/(\d[,.]\d)|([^\w\s](?!\s)(?!$))/$1 || "$2 "/ge;
[download]

Evaluating $1 || "$2 " means you get $1 if $1 is true (which means, anytime \d[.,]\d matches), otherwise you get $2 (the punctation character), followed by a space.

I would handle end-of-line processing as follows:

s/(\d[,.]\d)|([^\w\s])(?=\S)/$1 || "$2 "/ge;
[download]

A single positive look-ahead is, IMO, more clear than two negative look-aheads.

Abigail

$s =~ s/(?<=(?<!\d(?=[.,]\d))\p{P})(?!\s)/ /g;
[download]

The before condition is essentially (?<=\pP) with an additional condition. \pP matches any punctuation character. These are:

$ perl -wle'use charnames (); for (0..255) { chr($_) =~ /\pP/ and prin
+t chr($_)
," ($_): ", charnames::viacode($_) }'
! (33): EXCLAMATION MARK
" (34): QUOTATION MARK
# (35): NUMBER SIGN
% (37): PERCENT SIGN
& (38): AMPERSAND
' (39): APOSTROPHE
( (40): LEFT PARENTHESIS
) (41): RIGHT PARENTHESIS
* (42): ASTERISK
, (44): COMMA
- (45): HYPHEN-MINUS
. (46): FULL STOP
/ (47): SOLIDUS
: (58): COLON
; (59): SEMICOLON
? (63): QUESTION MARK
@ (64): COMMERCIAL AT
[ (91): LEFT SQUARE BRACKET
\ (92): REVERSE SOLIDUS
] (93): RIGHT SQUARE BRACKET
_ (95): LOW LINE
{ (123): LEFT CURLY BRACKET
} (125): RIGHT CURLY BRACKET
¡ (161): INVERTED EXCLAMATION MARK
« (171): LEFT-POINTING DOUBLE ANGLE QUOTATION MARK
· (183): MIDDLE DOT
» (187): RIGHT-POINTING DOUBLE ANGLE QUOTATION MARK
¿ (191): INVERTED QUESTION MARK
[download]

The additional condition is (?<!\d(?=[.,]\d)): this stipulates that what comes before the punctuation should not match \d(?=[.,]\d); that is a single digit, where the digit is followed by a period or comma and another digit. Note that that lookahead is actually escaping the bounds of the outer lookbehind, but that is perfectly ok to do.

Update: added actual character to punctuation table

Update: weird. backtick (GRAVE ACCENT) isn't considered punctuation

The best I can come up with is

#! /usr/bin/perl

use strict;

while( <DATA> ) {
    s/(?<!\d), *|, *(?=\D)/, /g;
    print;
}
__DATA__
1,2,a,b
1,a,2,b
1,a,b,2
a,1,2,b
__produces__
1,2, a, b
1, a, 2, b
1, a, b, 2
a, 1,2, b
[download]

But the use of alternation strikes me a bit as cheating. There is no doubt ABWTDI, but I can't see it for the moment.

#!/usr/local/bin/perl

use strict;

while (my $str = <DATA>) {
    chomp($str);
    $str =~ s/(\d+[.,]\d+       # Substitute a (simple) number ...
              |                 # or ...
              [^.,\d ]+)        # something without periods, comma's,
                                # digits or spaces ...
              ((?!$)[.,]?)      # followed by a period or comma, if
                                # possible, but not when at the end of
                                # the line ...
              /$1$2 /gx;        # by the number or string, followed by
                                # the found period or comma, and a
                                # trailing space
    print "<$str>\n";
}

__DATA__
1,2,a,b
a,1,b,2,
1,21,a,b,word,d.3,4.5.6

__END__
<1,2, a, b>
<a, 1,b, 2,>
<1,21, a, b, word, d. 3,4. 5.6>
[download]

In addition to the on-line documentation, you could check out Mastering Regular Expressions, 2nd Edition.

Arjen

I think most of the solutions so far will fail to put a space into 'x34,3b'.

There is also the question of what to do with '2,3,4,5'? Is that '2,3, 4,5' or '2, 3,4, 5' or '2, 3, 4, 5'? All are possible.

It will probably be very hard to use a single regex to do this if non-numeric bits can start and/or end with a digit.

I got close using s///e, but it was very messy.

use strict;
use warnings;

my $string = '1,2,a,b,3';

$string =~ s/( (?:(?<!\d)(?:,|\.)(?!\s*\d))     |
               (?:(?<=\d)(?:,|\.)(?=\s*)(?!\d)) |
               (?:(?<!\d)(?:,|\.)(?=\s*\d))     )
            /$1 /gx;

print $string, "\n";
[download]

It's really ugly, but triggers a substitution all of the following cases:

• Comma or dot is not surrounded by digits, with optional whitespace after the comma or dot.
• Comma or dot doesn't have a digit to the left, with optional whitespace after the comma or dot.
• Comma or dot doesn't have a digit to the right, with optional whitespace after the comma or dot.

So I think it meets your spec. If this is homework I would seriously recommend immersing yourself in the POD's so that you'll be able to explain to your professor how it is you came up with such a wild RE. For that matter, it's time that I re-read the POD's, as I'm not convinced that it really needs to be so explicit to work properly.

See perlre to learn about zero-width lookahead and lookbehind (positive and negative). Also read perlretut and perlop under the "Regexp quote like characters" section, for starters.

s/(?!\d[.,]\d)(.[^\w\s])(?!\s)/$1 /g;
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Hmm.. Just tried:

$_ = "1.2.a.b";
s/(?!\d[.,]\d)([^\w\s])(?!\s)/$1 /g;
print;
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Prints "1. 2. a. b"

--dda

s/(?<=\D)([,.]) *(?=\D)/$1 /g;
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perl -e '$a="1,23, 23, df,sd.asf,\"\"";$a=~s/(?<=\D)([,.]) *(?=\D)/$1 
+/g;print "$a\n";'
1,23, 23, df, sd. asf, ""
[download]